Question

Tg(arccos(-1/3)) Решите пожалуйста

Answer 1

Cos(arccos(-1/3))=-1/3
sin(arccos(-1/3))=sqrt(1-1/9)=(2/3)*sqrt(2)
tg(arccos(-1/3))=sin/c0s=-2*sqrt(2)

Answer 2

$\arccos( - \frac{1}{3} ) \in ( \frac{ \pi }{2} ; \pi )$ ;

$tg( \arccos( - \frac{1}{3} ) ) < 0$ ;

$tg^2 { \arccos( - \frac{1}{3} ) } = \frac{1}{ \cos^2{ \arccos( - \frac{1}{3} ) } } - 1 = \frac{1}{ ( - \frac{1}{3} )^2 } - 1 = \frac{1}{ 1/3^2 } - 1 = 9 - 1 = 8$ ;

$tg( \arccos( - \frac{1}{3} ) ) = - 2 \sqrt{2}$ ;